1. **Problem statement:** Given vectors \(\overrightarrow{AB} = -3i + 6j\) and \(\overrightarrow{AC} = 10i - 2j\) in triangle ABC, find: (a) The size of angle \(\angle BAC\) in degrees to 1 decimal place. (b) The exact area of triangle ABC. 2. **Find \(\angle BAC\):** The angle between two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is given by: $$\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}$$ Calculate the dot product: $$\overrightarrow{AB} \cdot \overrightarrow{AC} = (-3)(10) + (6)(-2) = -30 - 12 = -42$$ Calculate the magnitudes: $$|\overrightarrow{AB}| = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5}$$ $$|\overrightarrow{AC}| = \sqrt{10^2 + (-2)^2} = \sqrt{100 + 4} = \sqrt{104} = 2\sqrt{26}$$ Calculate \(\cos \theta\): $$\cos \theta = \frac{-42}{3\sqrt{5} \times 2\sqrt{26}} = \frac{-42}{6 \sqrt{130}} = \frac{-7}{\sqrt{130}}$$ Find \(\theta\): $$\theta = \cos^{-1} \left( \frac{-7}{\sqrt{130}} \right)$$ Using a calculator: $$\theta \approx 123.7^\circ$$ 3. **Find the exact area of \(\triangle ABC\):** Area is half the magnitude of the cross product of \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\): $$\text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}|$$ In 2D, the cross product magnitude is: $$|\overrightarrow{AB} \times \overrightarrow{AC}| = |x_1 y_2 - y_1 x_2|$$ Calculate: $$= |-3 \times (-2) - 6 \times 10| = |6 - 60| = |-54| = 54$$ Therefore: $$\text{Area} = \frac{1}{2} \times 54 = 27$$ **Final answers:** (a) \(\angle BAC \approx 123.7^\circ\) (b) Area of \(\triangle ABC = 27\)